hdu 3231 Box Relations(拓扑排序)

Box Relations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1108    Accepted Submission(s): 420
Special Judge

Problem Description

There are n boxes C₁, C₂, ..., C_n in 3D space. The edges of the boxes are parallel to the x, y or z-axis. We provide some relations of the boxes, and your task is to construct a set of boxes satisfying all these relations.

There are four kinds of relations (1 <= i,j <= n, i is different from j):

• I i j: The intersection volume of C_i and C_j is positive.
  
• X i j: The intersection volume is zero, and any point inside C_i has smaller x-coordinate than any point inside C_j.
  
• Y i j: The intersection volume is zero, and any point inside C_i has smaller y-coordinate than any point inside C_j.
  
• Z i j: The intersection volume is zero, and any point inside C_i has smaller z-coordinate than any point inside C_j.

.

 

Input

There will be at most 30 test cases. Each case begins with a line containing two integers n (1 <= n <= 1,000) and R (0 <= R <= 100,000), the number of boxes and the number of relations. Each of the following R lines describes a relation, written in the format above. The last test case is followed by n=R=0, which should not be processed.

 

Output

For each test case, print the case number and either the word POSSIBLE or IMPOSSIBLE. If it's possible to construct the set of boxes, the i-th line of the followingn lines contains six integers x1, y1, z1, x2, y2, z2, that means the i-th box is the set of points (x,y,z) satisfying x1 <= x <= x2, y1 <= y <= y2, z1 <= z <= z2. The absolute values of x1, y1, z1, x2, y2, z2 should not exceed 1,000,000.

Print a blank line after the output of each test case.

 

Sample Input

3 2 I 1 2 X 2 3 3 3 Z 1 2 Z 2 3 Z 3 1 1 0 0 0

 

Sample Output

Case 1: POSSIBLE 0 0 0 2 2 2 1 1 1 3 3 3 8 8 8 9 9 9 Case 2: IMPOSSIBLE Case 3: POSSIBLE 0 0 0 1 1 1

 

Source

2009 Asia Wuhan Regional Contest Hosted by Wuhan University

 

题意：三维座标下让你构造满足条件的n个长方体。有R个要求，I i j表示第i个长方体和第j个长方体必须有交集;   X  i  j 表示 i 任意一点的x座标必须小于 j 的;

            Y i j ,Z i j同理。

题解 拓扑排序。分别把三个面拆成三条线段，再拆成两个点，根据要求构造关系，再进行三次拓扑排序即可。

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<queue>
#include<set>
using namespace std;

const int N=2005;
const int M=2e5+5;

int n,m;
int c[N];
int topo[N],t;
vector<int>X[N],Y[N],Z[N];
int ans[3][N];

void init() {
    for(int i=0; i<=2*n; i++) {
        X[i].clear();
        Y[i].clear();
        Z[i].clear();
    }
    for(int i=0; i<n; i++) {
        X[i].push_back(i+n);
        Y[i].push_back(i+n);
        Z[i].push_back(i+n);
    }
}

bool dfs(int u,vector<int>G[]) {
    c[u]=-1;
    for(int v=0; v<G[u].size(); v++) {
        int to=G[u][v];
        if(c[to]<0)return false;
        else if(!c[to]&&!dfs(to,G))return false;
    }
    c[u]=1;
    topo[--t]=u;
    return true;
}

bool toposort(vector<int>G[]) {
    t=2*n;
    memset(c,0,sizeof (c));
    for(int u=0; u<2*n; u++) {
        if(!c[u])if(!dfs(u,G))return false;
    }
    return true;
}

bool solve() {
    if(!toposort(X))return false;
    for(int i=0; i<2*n; i++)
        ans[0][topo[i]]=i;
    if(!toposort(Y))return false;
    for(int i=0; i<2*n; i++)
        ans[1][topo[i]]=i;
    if(!toposort(Z))return false;
    for(int i=0; i<2*n; i++)
        ans[2][topo[i]]=i;
    return true;
}

int main() {
   // freopen("test.in","r",stdin);
    int ca=1;
    while(~scanf("%d%d",&n,&m)&&(n+m)) {
        char c[2];
        int x,y;
        init();
        while(m--) {
            scanf("%s%d%d",c,&x,&y);
            x--,y--;
            if(c[0]=='I') {
                X[x].push_back(y+n);
                X[y].push_back(x+n);
                Y[x].push_back(y+n);
                Y[y].push_back(x+n);
                Z[x].push_back(y+n);
                Z[y].push_back(x+n);
            } else if(c[0]=='X') {
                X[x+n].push_back(y);
            } else if(c[0]=='Y') {
                Y[n+x].push_back(y);
            } else {
                Z[n+x].push_back(y);
            }
        }
        if(!solve()) {
            printf("Case %d: IMPOSSIBLE\n\n",ca++);
            continue;
        }
        printf("Case %d: POSSIBLE\n",ca++);
        for(int i=0; i<n; i++) {
            printf("%d %d %d %d %d %d\n",ans[0][i],ans[1][i],ans[2][i],ans[0][i+n],ans[1][i+n],ans[2][i+n]);
        }
        putchar(10);
    }
    return 0;
}